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# A portion of $6600 is invested at a 5% annual return, while  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Senior Manager Joined: 30 Aug 2003 Posts: 325 Location: dallas , tx Followers: 1 Kudos [?]: 9 [0], given: 0 A portion of$6600 is invested at a 5% annual return, while [#permalink]  14 Nov 2003, 10:53
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A portion of $6600 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from the portion earning a 5% return is twice that of the other portion, what is the total income from the two investments after one year? (A)$180
(B) $270 (C)$300
(D) $320 (E)$360
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shubhangi

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[#permalink]  14 Nov 2003, 11:07
B.

I'll show how I got the answer after a couple people respond so they get a chance to work on it.
Manager
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270 is correct.. [#permalink]  14 Nov 2003, 11:41
x, 6600-x
5x = 2 * 3* (6600-x)
x = 3600

Income = (5/100)*3600 + (3/100)*(6600-3600)

= 180+90 = 270
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[#permalink]  14 Nov 2003, 23:28
hi shubhangi,
yeah the answer is 270 and here is how...........
take the amount of money invested at 5% as x and the mount invested at 3% as 6600-x . Going by what the sum says the following equation is formed 5x/100 = 2(3. 6600-x/100). we thus get the value of x by solving the equation to be 3600 which is the amount invested at 5%. the amount invested at 3%is thus 3000. just claculate the interest on these two amounts and you have your answer.

cheers
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[#permalink] 14 Nov 2003, 23:28
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